Question

Given: ΔABC is isosceles; AB ≅ AC

Prove: ∠B ≅ ∠C

Triangle A B C is shown. Sides B A and A C are congruent.

We are given that ΔABC is isosceles with AB ≅ AC. Using the definition of congruent line segments, we know that
AB = BC
.

Let’s assume that angles B and C are not congruent. Then one angle measure must be greater than the other. If m∠B is greater than m∠C, then AC is greater than AB by the
.

However, this contradicts the given information that
. Therefore,
, which is what we wished to prove.

Similarly, if m∠B is less than m∠C, we would reach the contradiction that AB > AC. Therefore, the angles must be congruent.

Answer 1

Fill in the blanks and correct the small error:

Given: ΔABC is isosceles; AB ≅ AC.
Prove: ∠B ≅ ∠C.

Using the definition of congruent line segments, we know that AB = AC. (not AB = BC)

Assume, for contradiction, that ∠B and ∠C are not congruent. Then one angle measure is greater than the other. Suppose m∠B > m∠C. Then the side opposite the larger angle is longer, so AC > AB by the theorem: "In a triangle, a larger angle is opposite a longer side."

However, this contradicts the given information that AB = AC. Therefore m∠B = m∠C, i.e., ∠B ≅ ∠C, which is what we wished to prove.

(If instead m∠B < m∠C we get the symmetric contradiction AB > AC.) Hence ∠B ≅ ∠C.